The method will invoke your lengthLongestWord() method from Part 1 to determine the longest word in the phrase. oIn a loop from i = 1 to the value of the longest length (int): oThe method will examine the length of each word. If the length equals i, it will concatenate that word into the String to be returned. oIt will do this for each successive word length. Don’t forget about the last word! oThe method will return back a String that has all the words arranged in order of their lengths, from smallest to largest (i = 1 to i = longest word in phrase’s length). See the example output for this method’s result down below!.

Tom is having a problem with his washing machine. he notices that the machine vibrates violently at a frequency of 1500 rpm due to an unknown rotating unbalance. the machine is mounted on 4 springs each having a stiffness of 10 kn/m. tom wishes to add an undamped vibration absorber attached by a spring under the machine the machine working frequency ranges between 800 rpm to 2000 rpm and its total mass while loaded is assumed to be 80 kg a) what should be the mass of the absorber added to the machine so that the natural frequency falls outside the working range? b) after a first trial of an absorber using a mass of 35 kg, the amplitude of the oscillation was found to be 10 cm. what is the value of the rotating unbalance? c) using me-3.5 kg.m, find the optimal absorber (by minimizing its mass). what would be the amplitude of the oscillation of the absorber?